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cuicus Terenud Hides Un Vived elements to be used: 26.982 137.327 Aluminum g/mol 112.411 Carbon Barium g/mol 12.011 Cadmium g

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Ans: 1. 3 Back ago t ALSO )3 (aq) 3 Basoa + 248 ༦༡༩༠༡) (S) 2. Molar ratio of Balz. Ale (SA): Balon Baloh : All 3 3 ! 2 & moleSo 3 mole Bach 3 mole Baso 7.59 X10*2 mole Baclą 7.59x102 mole Basoa So, amount of Barium Sulfate made theoretically is - 7.5So, mass that donot luat 3 315.153 glimbol 8.13 818-3 moble e noble 19. 3. grams. So, 19.3 grams do not reach. (N) % yield 92

So use the stoichiometric ratio concept to calculate the desired moles produced.

The ratio of barium chloride to aluminium sulfate is 3:1 so. The moles of aluminium sulfate consumed will be 1/3 of the moles of barium chloride.

So limiting reagent is barium chloride.

Find the left moles of aluminium sulfate. Then it's mass.

And from the moles of barium chloride calculate moles of barium sulfate formed. Then it's mass by multiplying with its molar mass.


answered by: ANURANJAN SARSAM
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